All categories

3 years ago

HELPPP!!! Pls /: need help asap

Mathematics

1 answer:

Llana [10]3 years ago

3 0

V(s) = s^3 should be the correct answer. Hope this Helps!

You might be interested in

Answer the question no.10

mr Goodwill [35]

If AB=AC then It is an isocoles triangle. Because of this, Angle ABC=ACB, meaning they are both 60°. This means that Angle BAC is also 60°. Becasue of circle theorms, the angle BOC=2BAC so angle BOC=120°

6 0

3 years ago

Help please will mark brainliest

anyanavicka [17]

Answer:

its is the first grap btw did u knopw u can just copy the quetion and paste it on the space bar above and presss enter it easier that way

Step-by-step explanation:

5 0

3 years ago

13.) Find the area of the shaded region if the side length of the square is 8 in.

sladkih [1.3K]

Answer:

<h3>Area of shaded region = 6.88 square inches</h3>

Step-by-step explanation:

Area of shaded region

When we subtract the area of the circle from the area of square, we get the area of the 4 corners. When we divide that by 2, we get the area of 2 corners(shaded).
$\sf \boxed{\text{Area of the shaded region =$\dfrac{Area \ of \ square - area \ of \ circle}{2}$}}$

<h3>Square: </h3>

Area of square = side *side

= 8 * 8

= 64 square inches

<h3>Circle:</h3>

diameter of the circle = side of the square

d = 8 in

r = 8/2 = 4 in

$\sf \text{Area of circle = \pi r^2}$ Area of circle = πr²

= 3.14 * 4 * 4

= 50.24 square inches

$\sf \text{Area of shade region = \dfrac{64 - 50.24 }{2}}$

$\sf \text{Area of the shade region = $\dfrac{64 - 50.24}{2}$}$

$\sf = \dfrac{13.76}{2}\\\\= 6.88 \ square inches$

8 0

2 years ago

PLEASE HELP RN

yulyashka [42]

Step-by-step explanation:

y = -9x - 45 -5

y = -9x - 50

slope = x coefficient = -9

we can use the point where

x = 0

y = -50

P1 (0, -50)

and when y = 0

-9x = 50

x = -50/9

P2 = (-50/9 , 0)

3 0

3 years ago

Read 2 more answers

Find y' if x = tan(y)

Travka [436]

Use the chain rule:

$x=\tan y$
$\dfrac{\mathrm d}{\mathrm dx}x=\dfrac{\mathrm d}{\mathrm dx}\tan y$
$1=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$
$1=\sec^2y\,y'$
$y'=\dfrac1{\sec^2y}=\cos^2y$

Furthermore, if you restrict the domain of $y$ , we could write

$x=\tan y\iff \tan^{-1}x=y$

and so we could also have

$y'=\cos^2(\tan^{-1}x)=\left(\dfrac1{\sqrt{x^2+1}}\right)^2=\dfrac1{x^2+1}$

5 0

4 years ago